3.614 \(\int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^3} \, dx\)

Optimal. Leaf size=166 \[ -\frac{316 a^2 \sqrt{a x+1} \sqrt{c-\frac{c}{a x}}}{15 \sqrt{1-a x}}+\frac{158 a^2 \sqrt{c-\frac{c}{a x}}}{15 \sqrt{1-a x} \sqrt{a x+1}}-\frac{2 \sqrt{c-\frac{c}{a x}}}{5 x^2 \sqrt{1-a x} \sqrt{a x+1}}+\frac{32 a \sqrt{c-\frac{c}{a x}}}{15 x \sqrt{1-a x} \sqrt{a x+1}} \]

[Out]

(158*a^2*Sqrt[c - c/(a*x)])/(15*Sqrt[1 - a*x]*Sqrt[1 + a*x]) - (2*Sqrt[c - c/(a*x)])/(5*x^2*Sqrt[1 - a*x]*Sqrt
[1 + a*x]) + (32*a*Sqrt[c - c/(a*x)])/(15*x*Sqrt[1 - a*x]*Sqrt[1 + a*x]) - (316*a^2*Sqrt[c - c/(a*x)]*Sqrt[1 +
a*x])/(15*Sqrt[1 - a*x])

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Rubi [A]  time = 0.233736, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6134, 6129, 89, 78, 45, 37} \[ -\frac{316 a^2 \sqrt{a x+1} \sqrt{c-\frac{c}{a x}}}{15 \sqrt{1-a x}}+\frac{158 a^2 \sqrt{c-\frac{c}{a x}}}{15 \sqrt{1-a x} \sqrt{a x+1}}-\frac{2 \sqrt{c-\frac{c}{a x}}}{5 x^2 \sqrt{1-a x} \sqrt{a x+1}}+\frac{32 a \sqrt{c-\frac{c}{a x}}}{15 x \sqrt{1-a x} \sqrt{a x+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c/(a*x)]/(E^(3*ArcTanh[a*x])*x^3),x]

[Out]

(158*a^2*Sqrt[c - c/(a*x)])/(15*Sqrt[1 - a*x]*Sqrt[1 + a*x]) - (2*Sqrt[c - c/(a*x)])/(5*x^2*Sqrt[1 - a*x]*Sqrt
[1 + a*x]) + (32*a*Sqrt[c - c/(a*x)])/(15*x*Sqrt[1 - a*x]*Sqrt[1 + a*x]) - (316*a^2*Sqrt[c - c/(a*x)]*Sqrt[1 +
a*x])/(15*Sqrt[1 - a*x])

Rule 6134

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Dist[(x^p*(c + d/x)^p)/(1 + (c*
x)/d)^p, Int[(u*(1 + (c*x)/d)^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*
d^2, 0] &&  !IntegerQ[p]

Rule 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^3} \, dx &=\frac{\left (\sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{1-a x}}{x^{7/2}} \, dx}{\sqrt{1-a x}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{(1-a x)^2}{x^{7/2} (1+a x)^{3/2}} \, dx}{\sqrt{1-a x}}\\ &=-\frac{2 \sqrt{c-\frac{c}{a x}}}{5 x^2 \sqrt{1-a x} \sqrt{1+a x}}+\frac{\left (2 \sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{-8 a+\frac{5 a^2 x}{2}}{x^{5/2} (1+a x)^{3/2}} \, dx}{5 \sqrt{1-a x}}\\ &=-\frac{2 \sqrt{c-\frac{c}{a x}}}{5 x^2 \sqrt{1-a x} \sqrt{1+a x}}+\frac{32 a \sqrt{c-\frac{c}{a x}}}{15 x \sqrt{1-a x} \sqrt{1+a x}}+\frac{\left (79 a^2 \sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{1}{x^{3/2} (1+a x)^{3/2}} \, dx}{15 \sqrt{1-a x}}\\ &=\frac{158 a^2 \sqrt{c-\frac{c}{a x}}}{15 \sqrt{1-a x} \sqrt{1+a x}}-\frac{2 \sqrt{c-\frac{c}{a x}}}{5 x^2 \sqrt{1-a x} \sqrt{1+a x}}+\frac{32 a \sqrt{c-\frac{c}{a x}}}{15 x \sqrt{1-a x} \sqrt{1+a x}}+\frac{\left (158 a^2 \sqrt{c-\frac{c}{a x}} \sqrt{x}\right ) \int \frac{1}{x^{3/2} \sqrt{1+a x}} \, dx}{15 \sqrt{1-a x}}\\ &=\frac{158 a^2 \sqrt{c-\frac{c}{a x}}}{15 \sqrt{1-a x} \sqrt{1+a x}}-\frac{2 \sqrt{c-\frac{c}{a x}}}{5 x^2 \sqrt{1-a x} \sqrt{1+a x}}+\frac{32 a \sqrt{c-\frac{c}{a x}}}{15 x \sqrt{1-a x} \sqrt{1+a x}}-\frac{316 a^2 \sqrt{c-\frac{c}{a x}} \sqrt{1+a x}}{15 \sqrt{1-a x}}\\ \end{align*}

Mathematica [A]  time = 0.0308127, size = 58, normalized size = 0.35 \[ -\frac{2 \left (158 a^3 x^3+79 a^2 x^2-16 a x+3\right ) \sqrt{c-\frac{c}{a x}}}{15 x^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - c/(a*x)]/(E^(3*ArcTanh[a*x])*x^3),x]

[Out]

(-2*Sqrt[c - c/(a*x)]*(3 - 16*a*x + 79*a^2*x^2 + 158*a^3*x^3))/(15*x^2*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.086, size = 69, normalized size = 0.4 \begin{align*} -{\frac{316\,{x}^{3}{a}^{3}+158\,{a}^{2}{x}^{2}-32\,ax+6}{15\, \left ( ax+1 \right ) ^{2}{x}^{2} \left ( ax-1 \right ) ^{2}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x)

[Out]

-2/15*(158*a^3*x^3+79*a^2*x^2-16*a*x+3)*(c*(a*x-1)/a/x)^(1/2)*(-a^2*x^2+1)^(3/2)/(a*x+1)^2/x^2/(a*x-1)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \sqrt{c - \frac{c}{a x}}}{{\left (a x + 1\right )}^{3} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*sqrt(c - c/(a*x))/((a*x + 1)^3*x^3), x)

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Fricas [A]  time = 2.1239, size = 142, normalized size = 0.86 \begin{align*} \frac{2 \,{\left (158 \, a^{3} x^{3} + 79 \, a^{2} x^{2} - 16 \, a x + 3\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a c x - c}{a x}}}{15 \,{\left (a^{2} x^{4} - x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorithm="fricas")

[Out]

2/15*(158*a^3*x^3 + 79*a^2*x^2 - 16*a*x + 3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x))/(a^2*x^4 - x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/x**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \sqrt{c - \frac{c}{a x}}}{{\left (a x + 1\right )}^{3} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*sqrt(c - c/(a*x))/((a*x + 1)^3*x^3), x)